Under all circumstances? No. If you immerse the circuit in a region with a changing magnetic field going through the circuit's loop, then Faraday's law tells you that the electric field circulation over the loop is proportional to the change in magnetic flux through the loop
$$
\oint_\mathcal C \mathbf E\cdot\mathrm d\mathbf l = -\frac{\mathrm d}{\mathrm dt}\iint_\mathcal S \mathbf B\cdot \mathrm d\mathbf S,
$$
where if your circuit consists only of resistive elements and cells then the integral on the left is the sum of the Ohm's-law voltages across the resistors and the cells' marked driving voltages.
In the case you posit, on the other hand, the situation is simpler in some ways. Here the Kirchhoff voltage law still holds, but what breaks is your assumptions, which for this situation are inconsistent. In particular, you can no longer say
negligible internal resistance
for either of the two cells, and you probably can't think of those resistances as linear circuit elements, either. Instead, you need to include the cells' internal resistance (however small) into the configuration, do the full Kirchhoff analysis, and then decide whether your cells are in their linear regime and whether the internal resistances are so small that removing them would not appreciably change the conclusions.
What you will find is that they're not removable, and you will likely be pushing charge through one of the batteries in reverse. Here the usual circuit abstractions break down: some voltage sources will accept this and keep their stride, but others can have nonlinear current-voltage characteristics, and many can sustain damage, from mild all the way up to catastrophic.
$Kirchhoff's\space law$ can prove useful here.
Case 1. Consider a closed circuit with two batteries, of emf $V_1$ and $V_2$, and a resistor of resistance R (all in series). A circuit with no resistance will catch fire. The resistor is added to prevent short circuiting.
According to Kirchhoff's law,
(i) If the $+ve$ terminal of a battery is connected to the $-ve$ of the other :
$V_1 + V_2 -IR=0$
In this arrangement of batteries, their emfs will just add up. No problem encountered here.
(ii) If the $+ve$ terminal of a battery is connected to the $+ve$ of the other :
$V_1 - V_2 -IR = 0$
In this arrangement of batteries, the emfs will subtract. The battery with larger emf will decide the current's direction. The current will be directed from the $-ve$ to the $+ve$ terminal of the battery with larger emf.
If batteries of same emf are used then their emfs will simply cancel out, resulting in zero current.
Case 2. In a circuit with parallel batteries as shown by the 2nd picture in your question, you can find out the direction of the current and the net emf just like it was found in Case 1. You just have to apply Kirchhoff's law correctly. Also, do not forget to insert a resistor.
Kirchhoff's law will prove to be a very useful tool. You just have choose a closed loop and use the sign conventions appropriately.
FYI, for kirchhoff's law to work you can assume any direction of current in the closed loop, either clockwise or anti-clockwise. If you assumed the correct direction of the current, you will get a positive value of current, otherwise negative implying the current is flowing in the direction opposite to the assumed one.
Best Answer
Real batteries have a finite energy storage capacity. Adding additional cells adds additional capacity (this is why I would add them, you haven't really specified any context so it's hard to say what you are looking for).
Also, it's worth noting:
This is an important assumption. In a real circuit, you would lose some power ($I^2R$) to the internal resistance of the battery. You can reduce this power loss by adding cells, reducing the current each cell provides.